A closed tank has two compartments A and B,bo | vedclass

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(A) Initial moles in compartment $A$ $(n_A)$: $n_A = \frac{P_A V_A}{R T_A} = \frac{5 	imes 1}{R 	imes 400} = \frac{1}{80R}$Initial moles in compartm

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$2.22$

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(A) Initial moles in compartment $A$ $(n_A)$: $n_A = \frac{P_A V_A}{R T_A} = \frac{5 	imes 1}{R 	imes 400} = \frac{1}{80R}$Initial moles in compartment $B$ $(n_B)$: $n_B = \frac{P_B V_B}{R T_B} = \frac{1 	imes 3}{R 	imes 300} = \frac{1}{100R}$Total moles $(n_{total})$ = $n_A + n_B = \frac{1}{80R} + \frac{1}{100R} = \frac{5+4}{400R} = \frac{9}{400R}$At equilibrium,the partition is conducting and movable,so both compartments will have the same pressure $(P)$ and temperature $(T)$.Let $V_A'$ and $V_B'$ be the new volumes. $V_A' + V_B' = 1 + 3 = 4 \ m^3$.Using the ideal gas law for the total system: $P(V_A' + V_B') = n_{total} R T$$P(4) = (\frac{9}{400R}) R T \implies P = \frac{9T}{1600}$For compartment $A$: $P V_A' = n_A R T \implies (\frac{9T}{1600}) V_A' = (\frac{1}{80R}) R T$$V_A' = \frac{1}{80} 	imes \frac{1600}{9} = \frac{20}{9} \approx 2.22 \ m^3$.

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